$2{e^ - }\, + \,2{H_2}O\, \to \,{H_2}\, + \,2O{H^ - }$

(valence factor) $H_2=2$

At $NTP$ $22400\,mL$ of $H_2=1\,mole$ of $H_2\,\,112\,mL$ of

${H_2}\, = \,\frac{1}{{22400}}\, \times \,112\, = \,0.005\,mole$ of $H_2$

Moles of $H_2$ produced

$ = \frac{{I\, \times t}}{{96500}}\, \times $ mole ratio

$0.05 = \frac{{I\, \times 965}}{{96500}}\, \times \frac{{1\,\,mole\,\,of\,\,{H_2}}}{{2\,\,mole\,\,of\,\,{e^ - }}}$

$I\,=\,1.0\,A$